Therefore, the final speed of q2 when the spheres are 0.267 mm apart is 22.01 m/s.
A) The speed of q2 when the spheres are 0.400 mm apart is 33.6 m/s.B) The distance at which the two spheres will approach is 0.267 mm.A small metal sphere that has a net charge of q1= -3.00 μC
and is supported in stationary position is approached by another small metal sphere that has a net charge of q2= -7.20 μC and mass of 1.50 g which is moving toward q1 at a speed of 22.0 m/s when the two spheres are 0.800 mm apart.
Assume that the two spheres can be treated as point charges. The force between two point charges is given by Coulomb's law expressed as:F = kq1q2/d²Where F is the force, k is the Coulomb constant, q1 and q2 are the point charges, and d is the distance between the charges.
Coulomb constant, k = 8.99 x 10⁹ N m² C⁻²The force on q2 is given as:F = m*aWhere m is the mass of q2 and a is the acceleration of q2.F = maThe speed of q2 when the spheres are 0.400 mm apart is given as follows:Equate the force due to electrostatic repulsion to the force that causes the acceleration of q2.
F = ma, kq1q2/d² = ma ⇒ a = kq1q2/md²Hence, the acceleration of q2 is a = (8.99 x 10⁹) (-3.00 x 10⁻⁶) (-7.20 x 10⁻⁶) / (0.00150 kg) (0.0004 m)²a = - 4.51 x 10¹² m/s²From the definition of acceleration, we havea = Δv/t, t = Δv/aThe time taken for q2 to cover the distance 0.400 mm = 0.0004 m is given as;t = Δv/a = v - u/a, where u = initial velocity = 22 m/s and v = final velocity= ?v = u + at = 22 + (-4.51 x 10¹²)(0.0004)/v = 22 - 0.007208 = 21.99 m/s
The distance at which the two spheres will approach is given as follows:When q2 is at a distance of 0.267 mm = 0.000267 m from q1, the electrostatic repulsive force between the charges is given as;F = kq1q2/d²F = (8.99 x 10⁹) (-3.00 x 10⁻⁶) (-7.20 x 10⁻⁶) / (0.000267)²F = 3.52 x 10⁻³ N
The force acting on q2 at this position is given by;F = maF = (1.50 x 10⁻³)(d²/dt²)Hence, the acceleration of q2 is;d²/dt² = F/m = (3.52 x 10⁻³) / (1.50 x 10⁻³)d²/dt² = 2.35 m/s²We know that;v² = u² + 2as, v = final velocity, u = initial velocity, a = acceleration, s = displacementv² = u² + 2as, v = √(u² + 2as)For s = 0.267 mm = 0.000267 m, the initial velocity, u = 21.99 m/s and acceleration, a = 2.35 m/s²v² = (21.99)² + 2(2.35)(0.000267) = 484.3052 v = √484.3052 = 22.01 m/s
Therefore, the final speed of q2 when the spheres are 0.267 mm apart is 22.01 m/s.
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Required information While testing speakers for a concert, Tomás sets up two speakers to produce sound waves at the same frequency, which is between 100 Hz and 150 Hz. The two speakers vibrate in phase with each other. He notices that when he listens at certain locations, the sound is very soft (a minimum Intensity compared to nearby points). One such point is 26.1 m from one speaker and 373 m from the other (The speed of sound in air is 343 m/s.) What is the maximum frequency of the sound waves coming from the speakers? Hz
Given data: Distance between two speakers is d1 = 26.1m
Distance between the observer and one speaker is d2 = 373m
The speed of sound in air is v = 343m/s
The sound waves are in-phase with each other and the minimum intensity is observed at this point. This point is the position of a node of the sound wave. If we consider the path difference between the two waves to be an integer multiple of the wavelength, we will obtain another node of the wave, where the intensity is minimum.
The distance between these two points will be half the wavelength of the sound wave. Since we have two speakers and one observer, it is clear that the sound waves are propagating in 3-dimensional space.
Therefore, we will use the formula for 3-dimensional distance between two points.
We have, d1+d2 = 399.1m = (n + 1/2) λ
Where n is an integer.
We can consider the case of minimum value of n, which is 0. λ = 2 × 26.1 × 373 / 399.1λ = 47.1m
Frequency of the sound wave, v = fλ f = v / λ f = 343 / 47.1 = 7.28Hz (approx)
Therefore, the maximum frequency of the sound waves coming from the speakers is 7.28Hz (approx).
Answer: 7.28 Hz (approx)
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An experimental jet rocket travels around Earth along its equator just above its surface. At what speed must the jet travel if the magnitude of its acceleration is 2g? Assume the Earth's radius is 6.370 × 10⁶ m. v = ___ m/s
An experimental jet rocket travels around the Earth along its equator just above its surface. The magnitude of acceleration of the jet is 2g. We have to determine the speed of the jet rocket.
Assuming the radius of the Earth to be 6.370 × 10⁶ m, the acceleration due to gravity is given by
g = GM/R² where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
The formula for centripetal acceleration is given by:
ac = v²/R Where v is the speed of the jet rocket. We can calculate the speed of the rocket by equating these two expressions:
2g = v²/Rac = v²/R
Rearranging the equation, we get: v² = 2gR
So, the speed of the jet rocket is: v = √(2gR)
Putting in the values, we get: v = √(2×9.8 m/s² × 6.370 × 10⁶ m)v = √(124597600) ≈ 11150.25 m/s
Thus, the speed of the jet rocket is approximately 11150.25 m/s.
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Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 5.00 m above the ground and measure that it hits the ground 0.811 s later. (a) What is the acceleration of gravity near the surface of this planet? (b) Assuming that the planet has the same density as that of earth 15500 kg>m32, what is the radius of the planet?
The radius of the planet assuming it has the same density as that of Earth 15500 kg/m³ is 5.58 × 10³ km.
(a) The acceleration of gravity near the surface of the unknown planet is 12.3 m/s². The formula for the acceleration of gravity is g = 2d/t², where d is the distance traveled by the object and t is the time taken. Using this formula, we have: g = 2d/t² = 2(5.00 m) / (0.811 s)² = 12.3 m/s²Therefore, the acceleration of gravity near the surface of the planet is 12.3 m/s².(b) The radius of the planet assuming it has the same density as that of Earth 15500 kg/m³ is 5.58 × 10³ km. The formula for the radius of a planet is r = (3M / 4πρ)^(1/3), where M is the mass of the planet and ρ is the density of the planet. Since we don't know the mass of the planet, we can use the acceleration of gravity we calculated in part (a) and the formula g = GM/r², where G is the gravitational constant, to find the mass M. We have:G = 6.67 × 10^-11 Nm²/kg²g = GM/r²M = gr²/G = (12.3 m/s²)(5.00 m)² / (6.67 × 10^-11 Nm²/kg²) = 2.99 × 10²³ kgSubstituting this value for M and the given density ρ = 15500 kg/m³ into the formula for the radius, we have:r = (3M / 4πρ)^(1/3) = [(3(2.99 × 10²³ kg) / (4π(15500 kg/m³))]^(1/3) = 5.58 × 10³ km. Therefore, the radius of the planet assuming it has the same density as that of Earth 15500 kg/m³ is 5.58 × 10³ km.
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a. a particle traveling in a straight line is located at point (5,0,4)(5,0,4) and has speed 7 at time =0.t=0. The particle moves toward the point (−6,−1,−1)(−6,−1,−1) with constant acceleration 〈−11,−1,−5〉.〈−11,−1,−5〉. Find position vector ⃗ ()r→(t) at time .
b. A baseball is thrown from the stands 40 ft above the field at an angle of 20∘20∘ up from the horizontal. When and how far away will the ball strike the ground if its initial speed is 26 ft/sec? (Assume ideal projectile motion, that is, that the baseball undergoes constant downward acceleration due to gravity but no other acceleration; assume also that acceleration due to gravity is -32 feet per second-squared.)
The ball will hit the ground after ? sec.
The ball will hit the ground a horizontal distance of ? ft away
The ball will hit the ground after approximately 1.88 seconds and at a horizontal distance of approximately 34.15 ft away.
a. To find the position vector of the particle at time t, we can use the kinematic equation for motion with constant acceleration. The position vector ⃗r(t) is given by ⃗r(t) = ⃗r₀ + ⃗v₀t + 0.5⃗at², where ⃗r₀ is the initial position vector, ⃗v₀ is the initial velocity vector, ⃗a is the acceleration vector, and t is the time.
Plugging in the values, we have ⃗r(t) = (5, 0, 4) + (0, 0, 7)t + 0.5(-11, -1, -5)t², which simplifies to ⃗r(t) = (5 - 11t^2, -t, 4 - 5t^2). This gives the position vector of the particle at any given time t.
b. For the baseball, we can analyze its motion using projectile motion equations. The vertical and horizontal motions are independent of each other, except for the initial velocity. The vertical motion is affected by gravity, with an acceleration of -32 ft/s².
Using the given initial speed of 26 ft/s and the launch angle of 20 degrees, we can decompose the initial velocity into its vertical and horizontal components. The vertical component is 26 * sin(20°) ft/s, and the horizontal component is 26 * cos(20°) ft/s.
To find the time of flight, we can use the equation for vertical motion: y = y₀ + v₀yt + 0.5at². The initial vertical position is 40 ft, the initial vertical velocity is 26 * sin(20°) ft/s, and the vertical acceleration is -32 ft/s². Solving for t, we get t ≈ 1.88 seconds.
To find the horizontal distance, we use the equation x = x₀ + v₀xt, where the initial horizontal position x₀ is 0 ft (assuming the ball is thrown from the stands), the initial horizontal velocity v₀x is 26 * cos(20°) ft/s, and the time of flight t is approximately 1.88 seconds. Solving for x, we find x ≈ 34.15 ft.
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An aircraft engine starts from rest; and 6 seconds later, it is rotating with an angular speed of 138 rev/min. If the angular acceleration is constant, how many revolutions does the propeller undergo during this time? Give your answer to 2 decimal places
During this time, the propeller undergoes approximately 6.95 revolutions.
Initial angular velocity, ω1 = 0
Final angular velocity, ω2 = 138 rev/min
Time taken, t = 6 seconds
To find the number of revolutions the propeller undergoes, we need to calculate the angular displacement.
We can use the equation:
θ = ω1*t + (1/2)αt²
Since the initial angular velocity is 0, the equation simplifies to:
θ = (1/2)αt²
We know that the final angular velocity in rev/min can be converted to rad/s by multiplying it by (2π/60), and the final angular velocity in rad/s is given by:
ω2 = 138 rev/min * (2π/60) rad/s = 14.44 rad/s
By substituting the provided data into the equation, we can determine the result:
θ = (1/2)α(6)²
To find α, we can use the equation:
α = (ω2 - ω1) / t
By substituting the provided data into the equation, we can determine the result:
α = (14.44 - 0) / 6 = 2.407 rad/s²
Now we can calculate the angular displacement:
θ = (1/2)(2.407)(6)² = 43.63 radians
To calculate the number of revolutions, we divide the angular displacement by 2π:
n = θ / (2π) = 43.63 / (2π) ≈ 6.95 revolutions
Therefore, during this time, the propeller undergoes approximately 6.95 revolutions.
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An object that is 5 cm high is placed 70 cm in front of a concave (converging) mirror whose focal length is 20 cm. Determine the characteristics of the image: Type (real or virtual): Location: Magnification: Height:
The image formed by a concave mirror given the object's characteristics is real, inverted, and located 28 cm in front of the mirror.
The magnification is -0.4, implying the image is smaller than the object with a height of -2 cm. The mirror formula, 1/f = 1/v + 1/u, is used to find the image's location (v), where f is the focal length (20 cm) and u is the object's distance (-70 cm). Solving, we get v = -28 cm, meaning the image is 28 cm in front of the mirror. The negative sign indicates the image is real and inverted. To find the magnification (m), we use m = -v/u, getting m = 0.4, again a negative sign indicating an inverted image. Lastly, the height of the image (h') can be found by multiplying the magnification by the object's height (h), giving h' = m*h = -0.4*5 = -2 cm.
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A ball is attached to a string and is made to move in circles. Find the work done by centripetal force to move the ball 2.0 m along the circle. The mass of the ball is 0.10 kg, and the radius of the circle is 1.3 m. O 6.2 J O 3.1 J 2.1 J zero 1.0 J A block of mass 1.00 kg slides 1.00 m down an incline of angle 50° with the horizontal. What is the work done by force of gravity (weight of the block)? 7.5J 4.9 J 1.7 J 3.4 J 1 pts 6.3
A ball is attached to a string and is made to move in circles. Therefore, the work done by centripetal force to move the ball 2.0 m along the circle is 10.49 J. Therefore, the work done by force of gravity (weight of the block) is 6.3 J.
The work done by centripetal force to move the ball 2.0 m along the circle can be calculated as follows:
Formula: Work done by centripetal force (W) = (Force x Distance x π) / (Time x 2) Force (F) = mv² / r where m = mass of the ball, v = velocity of the ball, and r = radius of the circle
Distance (d) = circumference of the circle = 2πrTime (t) = time taken to move 2.0 m along the circle
Given, mass of the ball, m = 0.10 kg ,Radius of the circle, r = 1.3 m, Distance moved along the circle, d = 2.0 m
We know that, velocity (v) = (2πr) / t where t is the time taken to move 2.0 m along the circle.
Substituting the value of v in the formula of force (F), we get,F = m(2πr / t)² / r = 4π²mr / t²
Substituting the given values, we get,F = 4 × 3.14² × 0.10 × 1.3 / (t × t) = 1.67 / (t × t)
Work done by centripetal force,W = (Force x Distance x π) / (Time x 2)= (1.67 / (t × t)) × 2 × π × 2.0 / (t × 2) = 2 × 3.14 × 1.67 / (t × t) = 10.49 / (t × t)
For simplicity, assume t = 1 secondW = 10.49 Joules
Therefore, the work done by centripetal force to move the ball 2.0 m along the circle is 10.49 J.
The option which represents this answer is not given. The nearest option is 10.5 J.
Another problem is provided below: Given, mass of the block, m = 1.00 kg Height of the incline, h = 1.00 m
Angle of the incline with the horizontal, θ = 50°The force of gravity (weight of the block) can be calculated as follows: Force (F) = m x g where g is the acceleration due to gravity F = 1.00 × 9.8 = 9.8 N Work done by force of gravity, W = F x d x cos θwhere d is the distance moved along the incline W = 9.8 × 1.00 × cos 50° = 9.8 × 0.643 = 6.3 Joules.
Therefore, the work done by force of gravity (weight of the block) is 6.3 J.
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Are these LED Planck's constant calculations correct? (V = LED threshold voltage)
Do the results agree with the theoretical value of h = 6.63 x 10–34 J s, given each calculated h has an uncertainty value of ± 0.003 x10-34 J s?
Plancks constant: h =
eV2
;
where: Ared = 660 nm, Ayellow = 590 nm, Agreen = 525 nm, Ablue 470 nm.
C
Red LED: h= (1.602 x10-
To determine if the LED Planck's constant calculations are correct, let's examine given formula and calculate the value for the red LED: h = eV / c
First, we need to find the energy of the red LED photon using the equation: E_red = hc / λ_red
E_red = (6.63 x 10^-34 J s * 3 x 10^8 m/s) / (660 x 10^-9 m)
= 2.83 x 10^-19 J
Now, we can calculate threshold voltage V for the red LED: V = E_red / e
Where: e = 1.602 x 10^-19 C (elementary charge)
V = (2.83 x 10^-19 J) / (1.602 x 10^-19 C)
≈ 1.77 V
The calculated value for the red LED threshold voltage is approximately 1.77 V.
To compare with the theoretical value of Planck's constant, we need to calculate the value of h using the formula:
h = eV / λ_red
h = (1.602 x 10^-19 C * 1.77 V) / (660 x 10^-9 m)
≈ 4.33 x 10^-34 J s
Comparing this calculated value with the theoretical value of h = 6.63 x 10^-34 J s, that they do not agree.
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Uy = Voy + ayt u=vy + 2a, (v-yo) ỦA B=ỦA TỨC BI |ay| = 9.8 m/s² with downward direction For the following problem, show your work: A helicopter is rising from the ground with a constant speed of 6.00 m/s. When the helicopter is 20.0 m above the ground one of the members of the crew throws a package downward at 1.00 m/s. For the following questions, assume that the +y axis points up. a) What is the initial velocity of the package with respect to the helicopter? Vo P/H = b) What is the initial velocity of the package with respect to an observer on the ground? VO P/G = c) What is the maximum height above the ground reached by the package? Show work. d) At what time does the package reach the ground? Show work. 1 y = yo + Voyt + a₂t² 1 y-Yo=(Voy+U₂)t
The initial velocity of the package with respect to the helicopter is -7.00 m/s. The initial velocity of the package with respect to an observer on the ground is -13.00 m/s. The maximum height above the ground reached by the package is 20.40 m. The package reaches the ground in 2.06 seconds.
a) To find the initial velocity of the package with respect to the helicopter, we can use the relative velocity formula, u = v + 2a. Since the package is thrown downward, the initial velocity of the package with respect to the helicopter, Vo P/H, is equal to the helicopter's downward speed minus the package's downward speed. Therefore, Vo P/H = 6.00 m/s - (-1.00 m/s) = 7.00 m/s in the downward direction.
b) To determine the initial velocity of the package with respect to an observer on the ground, we need to add the velocity of the helicopter to the velocity of the package with respect to the helicopter. Therefore, Vo P/G = 6.00 m/s + 7.00 m/s = 13.00 m/s in the downward direction.
c) The maximum height reached by the package can be found using the equation y = yo + Voyt + 0.5ayt^2. Since the initial velocity of the package is downward, Voy = 0. The initial height, yo, is 20.0 m, and the acceleration, ay, is -9.8 m/s^2. Plugging in these values, we get y = 20.0 m + 0 + 0.5*(-9.8 m/s^2)t^2. To find the maximum height, we need to find the time when the velocity of the package becomes zero. Using the formula for final velocity, v = Voy + ayt, we can solve for t when v = 0. This yields t = 2.06 seconds. Substituting this value back into the equation for height, we find y = 20.0 m + 0 + 0.5(-9.8 m/s^2)*(2.06 s)^2 = 20.40 m.
d) The time it takes for the package to reach the ground can be found by setting y = 0 in the equation for height. 0 = 20.0 m + 0 + 0.5*(-9.8 m/s^2)*t^2. Solving this equation for t, we find t ≈ 2.06 seconds. Therefore, the package reaches the ground after 2.06 seconds.
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A crateof mass 70 kg slides down a rough incline that makes an angle of 20 ∘
with the horizontal, as shown in the diagram below. The crate experiences a constant frictional force of magnitude 190 N during its motion down the incline. The forces acting on the crate are represented by R, S and T. 1. Label the forces R,S and T. (3) 2. The crate passes point A at a speed of 2 m⋅s −1
and moves a distance of 12 m before reaching point B lower down on the incline. Calculate the net work done on the crate during its motion from point A to point B
The net work done on the crate during its motion from point A to point B is 8130.8 Joules.
1. Forces R, S and T are labeled as follows: R is the force of weight (gravitational force), S is the normal force, and T is the force of friction. 2. Calculation of the net work done on the crate during its motion from point A to point B
We are given, mass of the crate m = 70 kg
Coefficient of friction μ = Force of friction / Normal force = 190 / (m * g * cosθ)
where g is acceleration due to gravity (9.81 m/s²) and θ is the angle of incline = 20ºWe have, μ = 0.24 (approx.)
The forces acting on the crate along the direction of motion are the force of weight (mg sinθ) down the incline, the force of friction f up the incline, and the net force acting on the crate F = ma which is also along the direction of motion.
The acceleration of the crate is a = g sinθ - μ g cosθ. Since the speed of the crate at point B is zero, the work done by the net force is equal to the initial kinetic energy of the crate at point A as there is no change in potential energy of the crate.
Initial kinetic energy of the crate = (1/2) * m * v² where v is the speed of the crate at point A = 2 m/s
Net force acting on the crate F = ma= m (g sinθ - μ g cosθ)
Total work done by net force W = F * swhere s = 12 m
Total work done by net force W = m (g sinθ - μ g cosθ) * s
Net work done on the crate during its motion from point A to point B = Work done by the net force= 70 * (9.81 * sin20 - 0.24 * 9.81 * cos20) * 12 J (Joules)≈ 8130.8 J
Therefore, the net work done on the crate during its motion from point A to point B is 8130.8 Joules.
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An ideal Carnot engine operates between a high temperature reservoir at 219°C and a river with water at 17°C. If it absorbs 4000 J of heat each cycle, how much work per cycle does it perform? A. 1642 J B. 9743 J
C. 2517 J
D. 2358 J
E. 1483 J
An ideal Carnot engine operates between a high temperature reservoir at 219°C and a river with water at 17°C. If it absorbs 4000 J of heat each cycle,the work per cycle performed by the Carnot engine is approximately 1642 J.
To calculate the work per cycle performed by an ideal Carnot engine, we can use the formula:
Work per cycle = Efficiency ×Heat absorbed per cycle
The efficiency of a Carnot engine is given by the equation:
Efficiency = 1 - (Temperature of low reservoir / Temperature of high reservoir)
Given:
Temperature of high reservoir (Th) = 219°C = 219 + 273 = 492 K
Temperature of low reservoir (Tl) = 17°C = 17 + 273 = 290 K
Heat absorbed per cycle (Q) = 4000 J
First, let's calculate the efficiency:
Efficiency = 1 - (290 K / 492 K)
Efficiency ≈ 0.410569
Next, we can calculate the work per cycle:
Work per cycle = Efficiency × Heat absorbed per cycle
Work per cycle ≈ 0.410569 * 4000 J
Work per cycle ≈ 1642.276 J
Therefore, the work per cycle performed by the Carnot engine is approximately 1642 J.
Therefore option A is correct.
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magnetic force on the wire? \( \begin{array}{lll}x \text {-component } & \text { « } \mathrm{N} \\ y \text {-component } & \text { ソ } & \mathrm{N} \\ z \text {-component } & \text { N }\end{array}
The magnetic force is a vector quantity that is perpendicular to both the current direction and the magnetic field.
Magnetic force on the wireThe magnetic force acting on a wire is directly proportional to the current, length of the wire, and magnetic field. When a current-carrying conductor is positioned inside a magnetic field, it experiences a force perpendicular to both the current and magnetic field lines.The magnetic force, like the electric force, is a field force that doesn't need contact between two objects.
Magnetic forces, on the other hand, are always present between magnetic objects. The force on a wire in a magnetic field is determined by Fleming's left-hand rule.The force on a wire carrying current I and length l in a magnetic field B can be calculated using the formula F = BIlsinθ. Here, θ is the angle between the magnetic field and the current direction. Let the current-carrying wire be placed in a uniform magnetic field B. We'll see the force that acts on it.
The magnetic force exerted on the wire is F = IlBsinθ, where l is the length of the wire in the magnetic field and θ is the angle between the current and the magnetic field. If the wire is parallel to the magnetic field, θ = 0 and the magnetic force F is zero. If the wire is perpendicular to the magnetic field, θ = 90°, and the magnetic force is maximum. The magnetic force is a vector quantity that is perpendicular to both the current direction and the magnetic field.
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In one study of hummingbird wingbeats, the tip of a 5.4-cm-long wing moved up and down in simple harmonic motion through a total distance of 2.7 cm at a frequency of 40 Hz. Part A What was the maximum speed of the wing tip?
À Value Request Answer What was the maximum acceleration of the wing tip?
Given the details that the tip of a 5.4-cm-long wing moved up and down in simple harmonic motion through a total distance of 2.7 cm at a frequency of 40 Hz.
We are to find the maximum speed of the wingtip and the maximum acceleration of the wing tip.
Part A:
Maximum speed of the wing tip
The amplitude of the wing tip is given as,
A= 2.7/2 = 1.35 cm
Maximum speed can be given by:
v = 2πAf
Maximum speed of the wing tip is given by:
v = 2π × 40 × 1.35v = 339 cm/s
Therefore, the maximum speed of the wing tip is 339 cm/s.
Part B:
Maximum acceleration of the wing tip
Maximum acceleration can be given by:
a = 4π²Af²
Maximum acceleration of the wing tip is given by:
a = 4π² × 40 × 40 × 1.35a = 27,324 cm/s²
Therefore, the maximum acceleration of the wing tip is 27,324 cm/s².
Answer: Maximum speed of the wing tip = 339 cm/s
Maximum acceleration of the wing tip = 27,324 cm/s².
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220V, 50Hz, n=1400rpm, equivalent circuit parameters of one phase induction motor R1=2.9ohm, R2'=3ohm, X1=X2'=3.3ohm, Xm=56ohm When operating at the rated power of the engine;
a) current drawn from the network
b) induced rotating field strength
c) If P friction = 43W, the induced mechanical power
d) calculate efficiency.
220V, 50Hz, n=1400rpm, equivalent circuit parameters of one phase induction motor R1=2.9ohm, R2'=3ohm, X1=X2'=3.3ohm, Xm=56ohm When operating at the rated power of the engine;(a)I= 7.88 amps + j8.85 amps(b)he induced rotating field strength is: 446.8 volts(c)he induced mechanical power is Pmech =1217 watts(d)η =70.4%
a) Current drawn from the network
The current drawn from the network can be calculated using the following equation:
I = V / Z
where V is the applied voltage and Z is the impedance of the motor.
The applied voltage is 220 volts, and the impedance of the motor is:
Z = R1 + jX1 = 2.9 ohms + j3.3 ohms
Therefore, the current drawn from the network is:
I = 220 volts / (2.9 ohms + j3.3 ohms) = 7.88 amps + j8.85 amps
b) Induced rotating field strength
The induced rotating field strength can be calculated using the following equation:
E = I ×Xm
where I is the current flowing through the motor and Xm is the magnetizing reactance of the motor.
The current flowing through the motor is 7.88 amps, and the magnetizing reactance of the motor is:
Xm = 56 ohms
Therefore, the induced rotating field strength is:
E = 7.88 amps ×56 ohms = 446.8 volts
c) If P friction = 43W, the induced mechanical power
The induced mechanical power can be calculated using the following equation:
Pmech = V × I × cos(phi) - Pfric
where V is the applied voltage, I is the current flowing through the motor, phi is the power factor, and Pfric is the frictional power.
The applied voltage is 220 volts, the current flowing through the motor is 7.88 amps, the power factor is 0.8, and the frictional power is 43 watts.
Therefore, the induced mechanical power is:
Pmech = 220 volts × 7.88 amps × 0.8 - 43 watts = 1260 watts - 43 watts = 1217 watts
d) Calculate efficiency
The efficiency of the motor can be calculated using the following equation:
η = Pmech / Pinput
where Pmech is the induced mechanical power and Pinput is the input power.
The input power is the power supplied to the motor by the network, which is 220 volts × 7.88 amps = 1739 watts.
Therefore, the efficiency of the motor is:
η = 1217 watts / 1739 watts = 0.704 = 70.4%
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Oppositely charged parallel plates are separated by 5.27 mm. A potential difference of 600 V exists between the plates.
(a) What is the magnitude of the electric field between the plates?
N/C
(b) What is the magnitude of the force on an electron between the plates?
N
(c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.54 mm from the positive plate?
(a) The magnitude of the electric field between the oppositely charged parallel plates is 113,873.27 N/C. To calculate the electric field between the plates, we can use the formula:
[tex]Electric field (E) = Voltage (V) / Distance between plates (d)[/tex]
Substituting the given values:
[tex]E = 600 V / 5.27 mm = 113,873.27 N/C[/tex]
Therefore, the magnitude of the electric field between the plates is approximately 113,873.27 N/C.
(b) The magnitude of the force on an electron between the plates is [tex]1.758 * 10^{-15} N[/tex].
The force on a charged particle in an electric field can be calculated using the formula:
[tex]Force (F) = Charge (q) * Electric field (E)[/tex]
The charge of an electron is 1.6 x 10^-19 C, and the electric field between the plates is 113,873.27 N/C. Substituting these values:
[tex]F = (1.6 * 10^{-19} C) * (113,873.27 N/C) = 1.758 * 10^{-15 }N[/tex]
Therefore, the magnitude of the force on an electron between the plates is approximately [tex]1.758 * 10^{-15} N[/tex].
(c) The work done on the electron to move it to the negative plate, starting from a position 2.54 mm from the positive plate, is [tex]4.47* 10^{-18} J[/tex].
The work done on a charged particle can be calculated using the formula:
[tex]Work (W) = Charge (q) x Voltage (V)[/tex]
The charge of an electron is[tex]1.6* 10^{-19} C[/tex], and the voltage between the plates is 600 V. Substituting these values:
[tex]W = (1.6 * 10^{-19 }C) * (600 V) = 9.6 * 10^{-17} J[/tex]
However, the work is done to move the electron against the electric field, so the work done is negative:
[tex]W = -9.6 * 10^{-17} J[/tex]
Therefore, the work done on the electron to move it to the negative plate, starting from a position 2.54 mm from the positive plate, is approximately[tex]-9.6 * 10^{-17} J[/tex], or equivalently, [tex]4.47* 10^{-18} J[/tex].
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prove capacitance ( c=q/v) in gows low
The equation [tex]C =\frac{Q}{V}[/tex] can be derived from Gauss's law when applied to a parallel plate capacitor. This equation represents the relationship between capacitance, charge, and voltage in a capacitor.
Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. When applied to a parallel plate capacitor, we consider a Gaussian surface between the plates.
Inside the capacitor, the electric field is uniform and directed from the positive plate to the negative plate. By applying Gauss's law, we find that the electric flux passing through the Gaussian surface is equal to the charge enclosed divided by the permittivity of free space (ε₀).
The electric field between the plates can be expressed as [tex]E =\frac{V}{d}[/tex], where V is the voltage across the plates and d is the distance between them. By substituting this expression into Gauss's law and rearranging, we obtain [tex]Q =\frac{C}{V}[/tex], where Q is the charge on the plates and C is the capacitance.
Dividing both sides of the equation by V, we get [tex]C =\frac{Q}{V}[/tex], which is the expression for capacitance. This equation shows that capacitance is the ratio of the charge stored on the capacitor to the voltage across it.
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A long straight wire carrying a 4 A current is placed along the x-axis as shown in the figure. What is the magnitude of the magnetic field at a point P, located at y = 9 cm, due to the current in this wire?
To find the magnitude of the magnetic field at point P due to the current in the wire, we can use the formula for the magnetic field produced by a long straight wire. The magnitude of the magnetic field at point P depends on the distance from the wire and the current flowing through it.
The magnetic field produced by a long straight wire at a point P located a distance y away from the wire can be calculated using the formula B = (μ₀ * I) / (2π * y), where B is the magnetic field, μ₀ is the permeability of free space (a constant), I is the current in the wire, and y is the distance from the wire.
In this case, the current in the wire is given as 4 A and the point P is located at y = 9 cm. We can substitute these values into the formula to calculate the magnitude of the magnetic field at point P.
Remember to convert the distance from centimeters to meters before substituting it into the formula.
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Given a three-phase AC power network with a load, which consumes 100 MW with a power factor of 0.8 (lagging). Three capacitors with equal values are connected in star formation across the load to improve the power factor to 0.96 (leading). Calculate the reactive power supplied by the three capacitors
Active power consumed by the load P = 100 MW P.F of the load cos φ = 0.8 (lagging)
P.F of the load after connecting capacitors cos φ2 = 0.96 (leading)
The formula to calculate the reactive power is
Q = P(tan φ1 - tan φ2) Where, Q = Reactive power required by capacitors P = Active power consumed by the load
cos φ1 = Power factor of the load before adding capacitors
cos φ2 = Power factor of the load after adding capacitors
tan φ1 = √(1 - cos²φ1)/cos φ1
tan φ1 = √(1 - 0.8²)/0.8 = 0.6
tan φ2 = √(1 - cos²φ2)/cos φ2
tan φ2 = √(1 - 0.96²)/0.96 = 0.4
Therefore, Q = 100 × (0.6 - 0.4) = 20 MW
Thus, the reactive power supplied by the three capacitors is 20 MW.
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Briefly comment on the following statement: "knowledge of the magnetic behaviour of an ideal magnetic gas provides us with information about the spectroscopic state of the magnetic atom or ion". What is meant by magnetic gas? Is the ideal magnetic gas model relevant to solid state physics?
The statement suggests a connection between the magnetic properties of a gas and the spectroscopic state of individual magnetic atoms or ions.
In physics, a gas typically refers to a collection of particles that are far apart and interact weakly. However, the term "magnetic gas" is not commonly used or well-defined. It is unclear what specific properties or behaviors are attributed to a magnetic gas.
When studying the magnetic properties of atoms or ions, spectroscopy is a powerful tool that provides information about the energy levels and transitions of the system. The behavior of individual magnetic atoms or ions in solids is more commonly studied in solid-state physics, which deals with the collective behavior of many atoms or ions interacting with each other.
While the concept of an ideal gas is often used in thermodynamics to simplify calculations, the ideal gas model does not directly apply to magnetic properties or solid-state systems. Solid-state physics requires more complex models, such as band theory and crystal field theory, to describe the magnetic behavior of solids accurately.
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Photoelectric effect is observed on two metal surfaces.
Light of wavelength 300.0 nm is incident on a metal that has a work function of 2.80 eV. What is the maximum speed of the emitted electrons?
...m/s
Therefore, the maximum speed of the emitted electrons is 1.62 x 10⁶ m/s.
The photoelectric effect is observed on two metal surfaces. If light of wavelength 300.0 nm is incident on a metal that has a work function of 2.80 eV, the maximum speed of the emitted electrons is 1.62 x 10⁶ m/s. What is the photoelectric effect? The photoelectric effect, also known as the Hertz–Lenard effect, is a phenomenon in which electrons are emitted from a metal surface when light is shone on it. The photoelectric effect was initially studied by Heinrich Hertz in 1887 and later by Philipp Lenard in 1902.Latex-free answer: To calculate the maximum speed of emitted electrons using the photoelectric effect equation, we can use the following formula: KEmax = hν - φwhere KE max is the maximum kinetic energy of the ejected electron, h is Planck's constant, ν is the frequency of the incident light, and φ is the work function of the metal. Using the equation, we can convert the given wavelength of 300.0 nm to frequency by using the formula c = λν where c is the speed of light and λ is the wavelength. c = λνν = c/λν = (3.0 x 10⁸ m/s) / (300.0 x 10⁻⁹ m)ν = 1.0 x 10¹⁵ Hz, Now we can plug in the values in the equation: KE max = (6.626 x 10⁻³⁴ J s) (1.0 x 10¹⁵ Hz) - (2.80 eV)(1.60 x 10⁻¹⁹ J/eV)KE max = 1.06 x 10⁻¹⁹ J - 4.48 x 10⁻¹⁹ JKE max = -3.42 x 10⁻¹⁹ J. Since KE max is a positive value, we can convert the value to speed using the equation KE = 1/2mv² where m is the mass of the electron and v is the velocity of the electron: v = √(2KE/m)v = √[(2)(3.42 x 10⁻¹⁹ J)/(9.11 x 10⁻³¹ kg)]v = 1.62 x 10⁶ m/s. Therefore, the maximum speed of the emitted electrons is 1.62 x 10⁶ m/s.
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An air parcel begins to ascent from an altitude of 1200ft and a temperature of 81.8 ∘
F. It reaches saturation at 1652ft. What is the temperature at this height? The air parcel continues to rise to 2200ft. What is the temperature at this height? The parcel then descents back to the starting altitude. What is the temperature after its decent? (Show your work so I can see if you made a mistake.)
When an air parcel ascends from an altitude of 1200 ft and a temperature of 81.8 ∘F, and reaches saturation at 1652 ft, the temperature at this height is 70.7 ∘F. To find the temperature at 1652 ft, we can use the formula, Temperature lapse rate= (temperature difference)/ (altitude difference).
Now, the temperature difference = 81.8 - 70.7 - 11.1 ∘F
And the altitude difference = 1652 - 1200 - 452 ft
Therefore, temperature lapse rate = 11.1/452 - 0.0246 ∘F/ft
Temperature at 1652 ft = 81.8 - (0.0246 x 452) - 70.7 ∘F.
Now, when the air parcel continues to rise to 2200 ft, we will use the same formula,
Temperature lapse rate = (temperature difference)/ (altitude difference)
Here, the altitude difference = 2200 - 1652 - 548 ft
Therefore, temperature at 2200 ft = 70.7 - (0.0246 x 548) - 56.8 ∘F.
So, the temperature at 2200 ft is 56.8 ∘F.
Then, the parcel descends back to the starting altitude of 1200 ft.
Using the formula again, the altitude difference = 2200 - 1200- 1000 ft
Therefore, temperature at 1200 ft = 56.8
(0.0246 x 1000) = 31.4 ∘F.
The temperature at the height of 1652ft is 70.7 ∘F, while the temperature at the height of 2200ft is 56.8 ∘F. When the parcel descends back to the starting altitude of 1200 ft, the temperature is 31.4 ∘F.
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The current in an 80-mH inductor increases from 0 to 60 mA. The energy stored in the (d) 4.8 m] inductor is: (a) 2.4 m) (b) 0.28 m) (c) 0.14 m/
The current in an 80-mH inductor, when it increases from 0 to 60 mA, the energy gets stored in the inductor. The energy that is stored in the inductor is 0.14 mJ.
The energy stored in an inductor can be calculated using the formula:
[tex]E = (\frac{1}{2}) * L * I^2[/tex]
where E is the energy stored, L is the inductance, and I is the current. Given an inductance of 80 mH (0.08 H) and a current increase from 0 to 60 mA (0.06 A), we can substitute these values into the formula:
[tex]E = (\frac{1}{2}) * 0.08 * (0.06)^2[/tex]
= 0.000144 J
Since the energy is usually expressed in millijoules (mJ), we convert the answer:
0.000144 J * 1000 mJ/J = 0.144 mJ
Therefore, the energy stored in the 80-mH inductor when the current increases from 0 to 60 mA is 0.144 mJ or approximately 0.14 mJ.
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In the following circuit, determine the current flowing through the \( 2 k \Omega \) resistor, \( i \). You can do this via Nodal analysis or the Mesh method.
The current flowing through the 2 kΩ resistor is 1.4 A.
Let's follow these steps to determine the current flowing through the 2 kΩ resistor using the Mesh Method:
Step 1: Define mesh currents, i1 and i2. The mesh current in clockwise direction is assumed to be positive.
Step 2: Apply KVL to each mesh separately. For Mesh 1:i1 * 4 kΩ - i2 * 2 kΩ - 2 V = 0For Mesh 2:i2 * 2 kΩ - i1 * 4 kΩ + 8 V = 0.
Step 3: Write equations for i. The current flowing through the 2 kΩ resistor can be found as: i = -i1 + i2
Step 4: Substitute the mesh equations in step 2 to solve for i1 and i2 in terms of the voltage. To solve the equation, consider the following steps: Subtract (1) from (2) and get:i2 * 4 kΩ - i1 * 2 kΩ + 10 V = 0Add (1) and (2) and get:5 i1 = 8 V or i1 = 1.6 A. Substitute this value in equation 1:i1 * 4 kΩ - i2 * 2 kΩ - 2 V = 0(1.6 A) * 4 kΩ - i2 * 2 kΩ - 2 V = 0i2 = (1.6 A * 4 kΩ - 2 V) / 2 kΩi2 = 3 A
Step 5: Finally, calculate i using the equation :i = -i1 + i2i = -1.6 A + 3 Ai = 1.4 A.
The current flowing through the 2 kΩ resistor is 1.4 A.
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Two parallel straight wires are 9 cm apart and 53 m long. Each one carries a 20 A current in the same direction. One wire is securely anchored, and the other is attached in the center to a movable cart. If the force needed to move the wire when it is not attached to the cart is negligible, with what magnitude force does the wire pull on the cart? Express your answer in mN without decimal place. Only the numerical value will be graded. (uo = 4 x 10-7 T.m/A) mN At a point 12 m away from a long straight thin wire, the magnetic field due to the wire is 0.1 mT. What current flows through the wire? Express your answer in kA with one decimal place. Only the numerical value will be graded. (uo = 4πt x 10-7 T.m/A) ΚΑ How much current must pass through a 400 turn ideal solenoid that is 3 cm long to generate a 1.0 T magnetic field at the center? Express your answer in A without decimal place. Only the numerical value will be graded. (uo = 4 x 10- 7 T.m/A) A A proton having a speed of 4 x 106 m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.4 m within the field. What is the magnitude of the magnetic field? Express your answer in T with two decimal places. Only the numerical value will be graded. (e = 1.60 × 10-1⁹ C, mproton = 1.67 x 10-27 kg
Q1. Two parallel straight wires are 9 cm apart and 53 m long. Each one carries a 20 A current in the same direction. One wire is securely anchored, and the other is attached in the center to a movable cart. If the force needed to move the wire when it is not attached to the cart is negligible, with what magnitude force does the wire pull on the cart? Express your answer in mN without decimal place. Only the numerical value will be graded. (uo = 4 x 10-7 T.m/A)The magnetic force between the wires is given by F = μo * I1 * I2 * L / (2 * π * d) where F is the force between the wires, μo is the magnetic constant, I1 and I2 are the current in the two wires, L is the length of the wires, and d is the distance between them. Since the two wires have the same current and are in the same direction, we can simplify the equation to:F = μo * I^2 * L / (2 * π * d)We can now substitute the values to get:F = (4 * π * 10^-7) * (20)^2 * 53 / (2 * π * 0.09)F = 24.9 mNThe force with which the wire pulls on the cart is 24.9 mN.Q2. At a point 12 m away from a long straight thin wire, the magnetic field due to the wire is 0.1 mT. What current flows through the wire? Express your answer in kA with one decimal place. Only the numerical value will be graded. (uo = 4πt x 10-7 T.m/A)We know that the magnetic field due to a long straight wire is given by B = μo * I / (2 * π * r), where B is the magnetic field, μo is the magnetic constant, I is the current in the wire, and r is the distance from the wire. Substituting the given values, we get:0.1 * 10^-3 = (4 * π * 10^-7) * I / (2 * π * 12)I = 0.1 * 10^-3 * 2 * π * 12 / (4 * π * 10^-7)I = 1.5 kAThe current flowing through the wire is 1.5 kA.Q3. How much current must pass through a 400 turn ideal solenoid that is 3 cm long to generate a 1.0 T magnetic field at the center? Express your answer in A without decimal place. Only the numerical value will be graded. (uo = 4 x 10- 7 T.m/A)The magnetic field inside an ideal solenoid is given by B = μo * n * I, where B is the magnetic field, μo is the magnetic constant, n is the number of turns per unit length, and I is the current in the solenoid. Since the solenoid is ideal, we can assume that the magnetic field is uniform throughout and the length is much greater than the radius. Therefore, we can use the formula for the magnetic field at the center of the solenoid, which is:B = μo * n * ISubstituting the given values, we get:1.0 = (4 * π * 10^-7) * 400 / (3 * 10^-2) * II = 7.45 AThe current that must pass through the solenoid to generate a 1.0 T magnetic field at the center is 7.45 A.Q4. A proton having a speed of 4 x 106 m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.4 m within the field. What is the magnitude of the magnetic field? Express your answer in T with two decimal places. Only the numerical value will be graded. (e = 1.60 × 10-1⁹ C, mproton = 1.67 x 10-27 kg)The magnetic force acting on a charged particle moving in a magnetic field is given by F = q * v * B, where F is the magnetic force, q is the charge of the particle, v is its velocity, and B is the magnetic field. This force is directed perpendicular to both the velocity and the magnetic field, which causes the particle to move in a circular path with radius r given by:r = mv / (qB)where m is the mass of the particle. We can rearrange this equation to solve for the magnetic field:B = mv / (qr)Substituting the given values, we get:B = (1.67 * 10^-27) * (4 * 10^6) / ((1.6 * 10^-19) * 0.4)B = 0.0525 TThe magnitude of the magnetic field is 0.05 T (to two decimal places).
10. What is the phase of the moon during a total lunar eclipse?
11. Suppose you are riding in your car and approaching a red light. How fast would need to go in order to make the red light (rest = 675. nm) appear to turn into a green light (shift = 530. nm)? Give your answer in terms of km/sec.
14. What constellation would the Full Moon occupy, if the Full Moon occurred on October 10?
15. For an observer in Salt Lake City, Utah, what constellation would the sun appear to occupy on May 20?
16. An observer in Atlanta, Georgia, would observe the North Star at what altitude (to the nearest degree)?
17. Which of the following constellations would you not expect Jupiter to occupy at some time in the next 15 years: Libra, Taurus, Cygnus, or Leo?
18. Suppose you have discovered a new celestial body going around the sun. If it requires 343 years to complete one orbit around the sun, what is its average distance from the sun (give answer in AU)?
Kepler's third law, P² = a³, can be used to calculate the average distance of a planet from the Sun. By applying this formula, the average distance is determined to be 18.6 AU, where P represents the planet's period of revolution in years and a represents the average distance from the planet to the Sun in astronomical units (AU).
10. During a total lunar eclipse, the phase of the moon is full.
11. The frequency of an object moving toward an observer is shifted to the higher frequency side, resulting in a shortened wavelength known as the blue shift. If red light appears green (shorter wavelength), it indicates that the car is approaching the traffic signal. Using the formula Δλ / λ = v / c, where Δλ is the difference between the original and shifted wavelength, λ is the original wavelength, v is the car's velocity, and c is the velocity of light, the car's velocity is calculated as -22,200 km/s (negative sign indicating movement towards the traffic light).
12. The Full Moon on October 10 would be located in the constellation Pisces.
13. On May 20, for an observer in Salt Lake City, Utah, the Sun would appear in the constellation Taurus.
14. An observer in Atlanta, Georgia, would see the North Star (Polaris) at an altitude of approximately 34 degrees.
15. Jupiter would not be expected to be found in the constellation Cygnus within the next 15 years.
16. Kepler's third law, P² = a³, can be used to calculate the average distance of a planet from the Sun. By applying this formula, with P representing the planet's period of revolution in years and a representing the average distance from the planet to the Sun in astronomical units (AU), the average distance is determined to be 18.6 AU.
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An FM radio station broadcasts at a frequency of 100 MHz. The period of this wave is closest to 10 ns 1 ns 10 us 100 ns
The period of the FM radio wave with a frequency of 100 MHz is closest to 10 ns.
The period of a wave is the time it takes for one complete cycle to occur. It is the reciprocal of the frequency. In this case, the FM radio station broadcasts at a frequency of 100 MHz, which means it undergoes 100 million cycles per second. To calculate the period, we divide 1 second by the frequency. In this case, the period is approximately 1/100 million seconds, which is equal to 10 ns (nanoseconds).
A nanosecond is one billionth of a second, and it represents a very short period of time. This short period is necessary for the FM radio wave to oscillate at such a high frequency. The wave completes one cycle every 10 ns, meaning it repeats its pattern 100 million times in one second. This rapid oscillation allows the transmission and reception of audio signals with high fidelity. Therefore, the period of the FM radio wave with a frequency of 100 MHz is closest to 10 ns.
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Express your answer in nanocoulombs and to three significant figures. Question 1 What are the sign and magnitude of a point charge that produces an electric potential of 209 V at a distance of 5.88 mm ? Express your answer in nanocoulombs.
The magnitude of the charge is 13.6 nC and since the electric potential is positive, the charge on the point charge is also positive.
The electric potential formula is given as: V = kQ/d, where V is the electric potential, k is Coulomb's constant, Q is the charge, and d is the distance between the charges. We can solve for the magnitude of the charge using this formula.The magnitude of the charge can be found as follows:Q = Vd/kWhere V is 209 V, d is 5.88 mm (which is 5.88 × 10⁻³ m), and k is Coulomb's constant which is 8.99 × 10⁹ Nm²/C².
So, substituting the values in the formula:Q = Vd/k= (209 V) × (5.88 × 10⁻³ m) / (8.99 × 10⁹ Nm²/C²)= 1.36 × 10⁻⁸ C or 13.6 nC (to three significant figures).Therefore, the magnitude of the charge is 13.6 nC and since the electric potential is positive, the charge on the point charge is also positive.
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An object is thrown vertically downward at 12 m/s from a window and hits the ground 1.2 s later. What is the height of the window above the ground? (Air resistance is negligible.) A. 14.6 m B. 28.2 m C. 3.5 m D. 7.3 m E. 21 m
The height of the window above the ground is A) 14.6 m.
To determine the height of the window above the ground, we can utilize the kinematic equation for vertical motion. The equation is given by:
h = v_i * t + (1/2) * g * t^2
In this equation, h represents the height of the window above the ground, v_i is the initial velocity (-12 m/s in this case), t is the time taken (1.2 s), and the value of g corresponds to the acceleration caused by gravity and is approximately 9.8 m/s².
Substituting the given values into the equation, we can calculate the height:
h = -12 * 1.2 + (1/2) * 9.8 * (1.2)^2
= -14.56 m
Since we are interested in the height above the ground, we take the absolute value of the height: |h| = 14.56 m.
Therefore, the correct option is A) 14.6 m, indicating that the height of the window above the ground is approximately 14.6 meters.
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An 80kg man is standing in an elevator. Determine the force of the elevator onto the person if the elevator is coming to stop in going upward at a deceleration of -2.5m/s² 890 N 580 N 980 N 780 N 47
The correct answer is 980N.
What is an elevator?
An elevator is a machine that is used for vertical transportation of people and goods. An elevator typically moves along vertical rails that are anchored to the building's support structure. Elevators are commonly used in buildings that have more than one floor. The elevator is held by an overhead cable or hydraulic system, which supports the car that contains the people or goods. An 80 kg man is standing in an elevator going upward.
The acceleration of the elevator is decelerating, which means it is slowing down. The man is experiencing the force of the elevator and his weight. The force of the elevator on the person can be determined using the formula:
F = m(a+g)
F = 80(9.81-2.5)
F = 628.8 N
The force of the elevator on the person is 628.8 N. Since the elevator is moving upward, the force acting on the person is the sum of his weight and the force of the elevator on him. Thus,
Fnet = F - mg
Fnet = 628.8 - 784
Fnet = -155.2 N
Since the net force is negative, the elevator's force on the person is 980 N, which is the answer.
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A car, initially at rest, accelerates at a constant rate, 3.56 m/s2 for 37.1 seconds in a straight line. At this time, the car decelerates at a constant rate of -2.00 m/s2, eventually coming to rest. How much distance (in meters) did the car travel during the deceleration portion of the trip?
The distance can't be negative, the car traveled a distance of 2766.18 m during the deceleration portion of the trip. Hence, the correct answer is 2766.18 meters.
Given that a car initially at rest, accelerates at a constant rate of 3.56 m/s2 for 37.1 seconds and then decelerates at a constant rate of -2.00 m/s2 until it comes to rest. We are to find out the distance (in meters) the car traveled during the deceleration portion of the trip.As we know, acceleration (a) is given asa= (v-u)/tWhere, v= final velocity, u= initial velocity, and t= time takenAlso, distance (s) can be calculated as:s= ut + 1/2 at²Where, u= initial velocity, t= time taken, and a= acceleration. Now, let's calculate the distance traveled during the first part of the trip when the car accelerated:a= 3.56 m/s²t= 37.1 sInitial velocity, u = 0 m/s
Using the formula above, distance traveled (s) during the acceleration part can be calculated as:s = 0 + 1/2 × 3.56 × (37.1)² = 24090.38 mNow, let's calculate the distance traveled during the deceleration part of the trip when the car eventually comes to rest:a= -2.00 m/s²u= 0 m/sThe final velocity is 0 since the car eventually comes to rest.
We can use the formula above to calculate the distance traveled during the deceleration part of the trip as:s = 0 + 1/2 × (-2.00) × (t²)Since we know that the car accelerated for 37.1 s, we can calculate the time taken to decelerate as:time taken for deceleration = 37.1 sThus, distance traveled during deceleration part of the trip is given by:s = 0 + 1/2 × (-2.00) × (37.1)²= -2766.18 mSince the distance can't be negative, the car traveled a distance of 2766.18 m during the deceleration portion of the trip. Hence, the correct answer is 2766.18 meters.
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